# Given f(x)=3x^4-2x^2 & g(x)= 2/sqrtx, (x ≠0) how do you find the composition of f and g?

Feb 2, 2018

See explanation.

#### Explanation:

There are two ways of composing 2 functions:

• $f \left(g \left(x\right)\right)$

To find this composition you have towrite the formula of $g \left(x\right)$ for every $x$ in the formula of $f$. Here we get:

$f \left(g \left(x\right)\right) = 3 \cdot {g}^{4} \left(x\right) - 2 \cdot {g}^{2} \left(x\right)$

$f \left(g \left(x\right)\right) = 3 \cdot {\left(\frac{2}{\sqrt{x}}\right)}^{4} - 2 \cdot {\left(\frac{2}{\sqrt{x}}\right)}^{2}$

$f \left(g \left(x\right)\right) = 3 \cdot \frac{16}{x} ^ 2 - 2 \cdot \frac{4}{x} = \frac{48}{x} ^ 2 - \frac{8}{x} = \frac{8 \cdot \left(6 - x\right)}{{x}^{2}}$

• $g \left(f \left(x\right)\right)$

To find this composition you have towrite the formula of $f \left(x\right)$ everywhere $x$ is in the formula of $g$. Here we get:

$g \left(f \left(x\right)\right) = \frac{2}{\sqrt{f \left(x\right)}} = \frac{2}{\sqrt{3 {x}^{4} + 2 {x}^{2}}}$